Answer: The value of x is -3. The value of y is 2. Step-by-step explanation: Given, 2x + y = -4 x + y = -1 To Find, The Value of x and y. Solution, 2x + y = -4 •••[1] x + y = -1 •••[2] Step 1 : Multiply Equation [2] by 2 to make the coefficients of x equal. Then we get the equations: 2x + y = -4 •••[3] 2x + 2y = -2 •••[4] Step 2 : Subtract Equation [3] from Equation [4] to eliminate y, because the coefficients of y are the same. So, we get: (2x + 2y) – (2x + y) = -2 – (-4) 2x + 2y – 2x – y = -2 + 4 y = 2 Step 3 : Substitute this value of y in [1], we get: 2x + y = -4 •••[1] 2x + 2 = -4 2x = -4 – 2 x = -6/2 x = -3 Verification, 2x + y = -4 •••[1] 2(-3) + 2 = -4 -6 + 2 = -4 -4 = -4 L.H.S. = R.H.S. x + y = -1 •••[2] -3 + 2 = -1 -1 = -1 L.H.S. = R.H.S. Required Answer, The value of x is -3. The value of y is 2. Reply
[tex]\boxed {\underline {\mathbb {FINAL\:ANSWER:-}}}[/tex] [tex]\boxed {x=-3}[/tex] [tex]\boxed {y=2}[/tex] [tex]\boxed {\underline {\mathbb {GIVEN:-}}}[/tex] [tex]2x+ y = -4\\x + y = -1[/tex] [tex]\boxed {\underline {\mathbb {TO\:FIND:-}}}[/tex] value of x and y [tex]\boxed {\underline {\mathbb {METHOD\:USED:-}}}[/tex] [tex]elimination \: method[/tex] [tex]\boxed {\underline {\mathbb {SOLUTION:-}}}[/tex] Let [tex]2x+y=-4…(1)[/tex] And [tex]x+y=-1…(2)[/tex] now by using eq2 let’s make eq3 we get- [tex]x+y=-1[/tex] [tex]x=-1-y[/tex] [ ← by bringing y to RHS] hence [tex]x=-1-y…(3)[/tex] now by using elimination method put eq3 in eq1 we get- [tex]2(-1-y)+y=-4[/tex] [ ←putted value in eq] [tex]-1 \times 2-y \times 2 + y =-4 \\-2-2y+y=-4\\-2-y=-4\\[/tex] [tex]-y=-4+2[/tex] [brought -2 to RHS get +2] [tex]-y=-2[/tex] [ ← when bring – to RHS we get – and – as +] [tex]\boxed {y=2}[/tex] as we get y value so let’s put in eq3 to get x value [tex]x=-1-(2)[/tex] [← putted y value in eq3] [tex]x=-1-2\\\boxed {x=-3}[/tex] ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ Reply
Answer:
The value of x is -3.
The value of y is 2.
Step-by-step explanation:
Given,
To Find,
Solution,
2x + y = -4 •••[1]
x + y = -1 •••[2]
Step 1 : Multiply Equation [2] by 2 to make the coefficients of x equal. Then we get the equations:
2x + y = -4 •••[3]
2x + 2y = -2 •••[4]
Step 2 : Subtract Equation [3] from Equation [4] to eliminate y, because the coefficients of y are the same. So, we get:
(2x + 2y) – (2x + y) = -2 – (-4)
2x + 2y – 2x – y = -2 + 4
y = 2
Step 3 : Substitute this value of y in [1], we get:
2x + y = -4 •••[1]
2x + 2 = -4
2x = -4 – 2
x = -6/2
x = -3
Verification,
2x + y = -4 •••[1]
2(-3) + 2 = -4
-6 + 2 = -4
-4 = -4
L.H.S. = R.H.S.
x + y = -1 •••[2]
-3 + 2 = -1
-1 = -1
L.H.S. = R.H.S.
Required Answer,
[tex]\boxed {\underline {\mathbb {FINAL\:ANSWER:-}}}[/tex]
[tex]\boxed {x=-3}[/tex]
[tex]\boxed {y=2}[/tex]
[tex]\boxed {\underline {\mathbb {GIVEN:-}}}[/tex]
[tex]2x+ y = -4\\x + y = -1[/tex]
[tex]\boxed {\underline {\mathbb {TO\:FIND:-}}}[/tex]
value of x and y
[tex]\boxed {\underline {\mathbb {METHOD\:USED:-}}}[/tex]
[tex]elimination \: method[/tex]
[tex]\boxed {\underline {\mathbb {SOLUTION:-}}}[/tex]
Let [tex]2x+y=-4…(1)[/tex]
And
[tex]x+y=-1…(2)[/tex]
now by using eq2 let’s make eq3
we get-
[tex]x+y=-1[/tex]
[tex]x=-1-y[/tex] [ ← by bringing y to RHS]
hence [tex]x=-1-y…(3)[/tex]
now by using elimination method put eq3 in eq1
we get-
[tex]2(-1-y)+y=-4[/tex] [ ←putted value in eq]
[tex]-1 \times 2-y \times 2 + y =-4 \\-2-2y+y=-4\\-2-y=-4\\[/tex]
[tex]-y=-4+2[/tex] [brought -2 to RHS get +2]
[tex]-y=-2[/tex] [ ← when bring – to RHS we get – and – as +]
[tex]\boxed {y=2}[/tex]
as we get y value so let’s put in eq3 to get x value
[tex]x=-1-(2)[/tex] [← putted y value in eq3]
[tex]x=-1-2\\\boxed {x=-3}[/tex]
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